Question

Can somebody help me with this Anti-Derivative problem??

Answer 1

do you know trig substitution?

Answer 2

i hope so

Answer 3

\[\frac{df}{dx}=\frac{3}{\sqrt{1-x^2}}\]\[f\left (\frac{1}{2}\right ) = 8.\]
\[df=\frac{3}{\sqrt{1-x^2}}dx\]\[f(x)=3\sin^{-1}(x)+c\]\[8=3\sin^{-1}\left(\frac{1}{2}\right )+c\]\[c=8-3\sin^{-1}\left(\frac{1}{2}\right )\]\[f(x)=\frac{3}{\sqrt{1-x^2}}+8-3\sin^{-1}\left(\frac{1}{2}\right )\]

Answer 4

\[\text{ Let } x=\cos(\theta) => dx=-\sin(\theta) d \theta\]
\[3\int\limits_{}^{}\frac{1}{\sqrt{1-\cos^2(\theta)}} -\sin(\theta) d \theta=3\int\limits_{}^{}-1 d \theta=-3 \theta+C\]

Answer 5

\[=-3\cos^{-1}(x)+C\]

Answer 6

I didnt know we needed trig substution.. My professor didnt cover this.

Answer 7

\[f(x)=-3 \cos^{-1}(x)+C\]

Answer 8

there is also a formula you can use

Answer 9

i don't do formulas

Answer 10

Oh ok.. Thank you

Answer 11

whoops, messed up above xd zzz

Answer 12

\[f(\frac{1}{2})=-3\cos^{-1}(\frac{1}{2})+C=8 =>-3 \cdot \frac{\pi}{3}+C=8\]
\[C=8+\pi\]

Answer 13

\[f(x)=-3 \cos^{-1}(x)+8+\pi\]

Answer 14

\[f(x)=3\sin^{-1}(x)-3\sin^{-1}\left (\frac{1}{2}\right )+8\]

Answer 15

-cos inverse
or sin inverse whatever!
same thing

Answer 16

Thank you myininaya! :)

Answer 17

\[f(x)=-3\cos^{-1}(x)+\pi+8\]

Answer 18

i could had made the subsition
x=sin(theta) instead and got the sin inverse

Answer 19

would you like to see?

Answer 20

Nah what you have done is good. Thank you!

Answer 21

ok have fun

Answer 22

oops one more thing
when i said thing i mean they just differ by a constant

Answer 23

samething*